The problem is:

Let $W$ be the set of all $(x_1,x_2,x_3,x_4,x_5)$ in $\mathbb{R}^5$ which satisfy

$$\begin{array}2x_1-x_2+\frac43x_3-x_4=0\\ x_1+\frac23x_3-x_5=0\\ 9x_1-3x_2+6x_3-3x_4-3x_5=0\end{array}$$

Find a finite set of vectors which spans $W$

The calculation shows its simplifies to row echelon form


And the solution says:

Thus, the system is equivalent to: $$x_1+\frac23x_3-x_5=0\\ x_2+x_4-2x_5=0$$

And it proceeds by saying that:

Thus the system is parametrized by $(x_3,x_4,x_5)$

Setting each equal to $1$ and the other $2$ equal to $0$, in turn, gives the three vectors $\begin{pmatrix}-\frac23&0&1&0&0\end{pmatrix}$, $\begin{pmatrix}0&-1&0&1&0\end{pmatrix}$, and $\begin{pmatrix}1&2&0&0&1\end{pmatrix}$, these three vectors therefore span $W$

I have totally no idea how did they determine the parametrizing variables and got the three vectors.

Thank you in advance.


2 Answers 2


Now, after getting the equivalence system, we can write that,

$$W = \{(x_1,x_2,x_3,x_4,x_5) \in \Bbb{R}^5 : x_1+\frac{2}{3}x_3-x_5=0, x_2+x_4-2x_5=0\} \\ = \{(x_5 - \frac{2}{3}x_3,2x_5 -x_4,x_3,x_4,x_5) : x_3,x_4,x_5 \in \Bbb R\} \\ = \{x_3(-\frac{2}{3},0,1,0,0) + x_4(0,-1,0,1,0) + x_5(1,2,0,0,1) : x_3,x_4,x_5 \in \Bbb R\}.$$

Thats why the three vectors $(-\frac{2}{3},0,1,0,0),(0,-1,0,1,0), (1,2,0,0,1)$, spans $W$.

  • 1
    $\begingroup$ Thank you very much! $\endgroup$
    – Yinuo An
    Commented Jul 10 at 10:56
  • "I have totally no idea how did they determine the parametrizing variables...": The non-parametrizing variables are the 1st and 2nd one because the 1st and 2nd columns contain a pivot. As a result, when you solve the system from the last line to the first, those variables will be expressed as linear combinations of the other ones: $$x_2=-x_4+2x_5,\quad x_1=-\frac23x_3+x_5,$$ hence so will be the whole generic solution: $$(x_1,\dots,x_5)=\left(-\frac23x_3+x_5,-x_4+2x_5,x_3,x_4,x_5\right).$$
  • "...and got the three vectors." The three vectors $$u:=\begin{pmatrix}-\frac23&0&1&0&0\end{pmatrix},v:=\begin{pmatrix}0&-1&0&1&0\end{pmatrix},w:=\begin{pmatrix}1&2&0&0&1\end{pmatrix}$$ were obtained by setting $(x_3,x_4,x_5)$ to (respectively) $(1,0,0),(0,1,0),(0,0,1),$ and you can check that the generic solution is then equal to $x_3u+x_4v+x_5w$.

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